A stone is thrown straight down from the edge of a roof, 825 ft above the ground, at a speed of 5 ft/sec.
A. Given that the acceleration due to gravity is -32 ft/sec2, how high is the stone 2 seconds later?
751 ft.
B. At what time does the stone hit the ground?
7.02615 sec.
C. What is the velocity of the stone when it hits the ground?
___ft/sec.
I can't find C
From your data
height = -16t^2 - 5t + 825
Agree with your a) and b) answers
v = -32t - 5
when t = 7.026 , v = -32(7.026) - 5
= ...... ft/s
I did that too and I got -299.823 but it is showing that it's wrong, idk why
I got -229.8384
since it asked for velocity, you must use the negative sign,
or else say 229.8364 ft/sec downwards or something along those lines
To find the velocity of the stone when it hits the ground, we need to find the time it takes for the stone to reach the ground first.
Let's break down the problem step by step:
1. The stone is thrown straight down from the edge of the roof, 825 ft above the ground, with an initial velocity of 5 ft/sec. Since the stone is thrown straight down, the initial velocity has a negative sign (-5 ft/sec).
2. The acceleration due to gravity is -32 ft/s^2, which means the stone's velocity is decreasing at a rate of 32 ft/sec every second.
A. To find the height of the stone 2 seconds later, we can use the kinematic equation:
h = h0 + v0t + (1/2)at^2
where:
h = height of the stone after time t
h0 = initial height (825 ft)
v0 = initial velocity (-5 ft/sec)
a = acceleration due to gravity (-32 ft/s^2)
t = time (2 seconds)
Plugging in the values, we have:
h = 825 + (-5)(2) + (1/2)(-32)(2)^2
= 825 - 10 - (1/2)(32)(4)
= 825 - 10 - 64
= 825 - 74
= 751 ft
Therefore, the height of the stone 2 seconds later is 751 ft.
B. To find the time it takes for the stone to hit the ground, we need to use the kinematic equation:
h = h0 + v0t + (1/2)at^2
Since the stone hits the ground, the final height h would be 0ft. We need to find the time t.
Rearranging the equation, we get:
(1/2)at^2 + v0t + h0 = 0
Plugging in the values:
(1/2)(-32)t^2 + (-5)t + 825 = 0
This is a quadratic equation. Solve for t using the quadratic formula or factoring. Once you solve it, you will find that t ≈ 7.02615 seconds.
Therefore, the stone hits the ground at approximately 7.02615 seconds.
C. To find the velocity of the stone when it hits the ground, we can use the kinematic equation:
v = v0 + at
where:
v = final velocity (what we are trying to find)
v0 = initial velocity (-5 ft/sec)
a = acceleration due to gravity (-32 ft/s^2)
t = time it takes for the stone to hit the ground (7.02615 seconds)
Plugging in the values, we have:
v = -5 + (-32)(7.02615)
Calculating the expression, we find that the velocity is approximately -225.63 ft/sec.
Therefore, the velocity of the stone when it hits the ground is approximately -225.63 ft/sec. The negative sign indicates that the velocity is directed downwards.